Problem: Solve for $x$ : $3\sqrt{x} - 6 = 9\sqrt{x} + 8$
Subtract $3\sqrt{x}$ from both sides: $(3\sqrt{x} - 6) - 3\sqrt{x} = (9\sqrt{x} + 8) - 3\sqrt{x}$ $-6 = 6\sqrt{x} + 8$ Subtract $8$ from both sides: $-6 - 8 = (6\sqrt{x} + 8) - 8$ $-14 = 6\sqrt{x}$ Divide both sides by $6$ $\frac{-14}{6} = \frac{6\sqrt{x}}{6}$ Simplify. $-\dfrac{7}{3} = \sqrt{x}$ The principal root of a number cannot be negative. So, there is no solution.